Re: calculating inductive factor for c cores

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Malcolm Watts by way of Terry Fritz <twftesla-at-qwest-dot-net>" <m.j.watts-at-massey.ac.nz>

Hi Justin,
            I can answer some questions for you:

On 6 Nov 2002, at 18:23, Tesla list wrote:

 > Original poster: "Justin Wright by way of Terry Fritz 
<twftesla-at-qwest-dot-net>" <justin-at-tracesofnut-dot-com>
 >
 > Hi All,
 >
 > I have some C Cores which I might like to rewind as HV transformers.
 >
 > How do I calculate the how much power a core can handle.

You can get a rough estimate from: 39 * Ae^2  where Ae is the
effective pole cross-sectional area. The derivation for this arises
from geometrical limitations on transformer design, in particular an
acceptable degree of regulation on the loaded secondary winding/s.
Although one can build a transformer core with a huge winding window
and relatively small pole area, the further the windings are from the
core, the poorer the coupling to the core will be and hence
regulation also suffers. Additionally, the windings will also suffer
from rather poor coupling between each other (not to mention between
their own turns).

  Can I get an
 > inductive factor from the core size and then work out the maximum
 > current it will allow?

Current throughput is a function of copper area of the winding wire
and the number of turns (the winding length) and by proxy, the
winding resistance and the ability of the windings to lose power
wasted as heat in that resistance. For a given copper mass, the
thicker the wire, the fewer the turns, the lower the terminal
voltage, the higher the current capability. A simple equation turns
this mess into a turns/volt figure for the core and if the primary
occupies around 40 - 45% of the total winding window, the magnetizing
current will be a low percentage of the reflected load current at the
core's VA rating. The equation can be boiled down to:  turns/volt =
7.164/Ae for the 60Hz mains.

 > Also, is it a good idea to have the primary and secondary separate from
 > each the, rather than the secondary over the top of the primary?

That depends on whether you want good coupling and regulation or a
high degree of leakage inductance. For good regulation (and fuse
blowing ability should a short be applied across the secondary), you
want the two windings to be as tightly coupled as possible
(interleaved does best but might not be suitable for HV secondaries).

 > Lastly, what is the approximate breakdown voltage of the enamel coating
 > on magnet wire?

It depends on the insulation type and goes up with wire thickness as
the coating thickness also increases in proportion. My preference is
polyesterimide and I have measured it withstanding some kV (about 7
from memory) between touching pieces of 1mm wire.

Regards,
malcolm